Cho \(\frac{a}{b}=\frac{c}{d}.\) Chứng minh:
\(a.\) \(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
\(b.\) \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
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1 tháng 11 2016
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\left(đpcm\right)\)
1 tháng 11 2016
\(\frac{a}{b}\) =\(\frac{c}{d}\) =>\(\frac{a}{c}\) =\(\frac{b}{d}\) =\(\frac{a-b}{c-d}\) =>\(\frac{ab}{cd}\) = \(\frac{a}{c}\) x\(\frac{b}{d}\) = \(\frac{a-b}{c-d}\) x \(\frac{a-b}{c-d}\) = \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Còn với\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) thì bạn chỉ cần thay dấu trừ thành dấu công là được
Chúc bạn học tốt
S
1 tháng 10 2017
1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
KD
20 tháng 10 2016
Cho \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=b.k;c=d.k\)
Vế trái:
\(\frac{a}{b}=\frac{c}{d}=\frac{b.k.b}{d.k.d}=\frac{b^2}{d^2}\)(1)
Vế phải:
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right).2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2) ta có:
\(\frac{ab}{c\text{d}}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(đpcm)
20 tháng 10 2016
ta có \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{a}=\frac{a+b}{c+d}\\\Rightarrow\frac{a}{c}.\frac{a}{c}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\\ \Rightarrow\frac{a}{c}.\frac{b}{d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}hay\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2} \)
TH
26 tháng 7 2016
a) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{a.a}{c.c}=\frac{b.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{a}{c}.\frac{b}{d}=\frac{a-b}{c-d}.\frac{a-b}{c-d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
HT
23 tháng 9 2015
Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Đpcm)
Có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
=> \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)(Đpcm)
30 tháng 7 2015
a,đặt a/b=c/d=k=>a=bk;c=dk khi đó ta có
ab/cd=bkb/dkd=b2k/d2k=b2/d2
a2-b2/c2-d2=b2k2-b2/d2k2-d2=b2(k2-1)/d2(k2-1)=b2/d2
=>ab/cd=a2-b2/c2-d2
L
17 tháng 1 2020
\(a)\) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có:
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(1\right)\)
Lại có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\left(đpcm\right)\)
\(b)\) Đặt \(\frac{a}{b}=\frac{c}{k}=k\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có:
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\frac{b^2}{d^2}\left(1\right)\)
Lại có:
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\left(đpcm\right)\)
26 tháng 12 2019
a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2), ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\frac{\left[b.\left(k-1\right)\right]^2}{\left[d.\left(k-1\right)\right]^2}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2), ta có: \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
CC
26 tháng 12 2019
a) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)
LA
7 tháng 3 2018
Ta có: \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Lại có: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{kb^2}{kd^2}=\frac{b^2}{d^2}\)
Tương tự: \(\frac{a^2+b^2}{c^2+d^2}=\frac{k^2b^2+b^2}{k^2d^2+d^2}=\frac{b^2\left(k+1\right)}{d^2\left(k+1\right)}=\frac{b^2}{d^2}\)
=> đpcm
Giải:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\)
a, Ta có: \(k^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\) (1)
\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
b, Ta có: \(k=\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow k^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (1)
\(k^2=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)Thay vào \(\frac{a^2-b^2}{c^2-d^2}\) ta được:
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{b^2k^2-b^2}{d^2k^2-d^2}\Rightarrow\frac{b^2}{d^2}\Rightarrow\frac{b.b}{d.d}\left(1\right)\)
Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow a=b;c=d\left(2\right)\)
Từ (1) và (2) suy ra:\(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)